Number of k consecutive 1's subsequences in a binary string.












0












$begingroup$




Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?



Example: in the sequence $1110011$ we have $N_2=3.$








probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
    $endgroup$
    – lulu
    Dec 19 '18 at 12:13












  • $begingroup$
    If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
    $endgroup$
    – player100
    Dec 19 '18 at 12:24










  • $begingroup$
    It does not have to be exactly k,I added an example for clearance.@player100
    $endgroup$
    – Adddison
    Dec 19 '18 at 12:29
















0












$begingroup$




Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?



Example: in the sequence $1110011$ we have $N_2=3.$








probability expected-value







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
    $endgroup$
    – lulu
    Dec 19 '18 at 12:13












  • $begingroup$
    If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
    $endgroup$
    – player100
    Dec 19 '18 at 12:24










  • $begingroup$
    It does not have to be exactly k,I added an example for clearance.@player100
    $endgroup$
    – Adddison
    Dec 19 '18 at 12:29














0












0








0





$begingroup$




Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?



Example: in the sequence $1110011$ we have $N_2=3.$








probability expected-value







share|cite|improve this question











$endgroup$






Say a sequence ${X_1, X_2,ldots ,X_n}$ is given, where $X_p$ is either one or zero ($0 < p < n$) with
probability $frac{1}{2}$ each. Let $N_k$ be the number of consecutive 1's subsequences of length k in ${X_1, X_2,ldots ,X_n}$,what is $Bbb E[N_k]$?



Example: in the sequence $1110011$ we have $N_2=3.$








probability expected-value




probability expected-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 12:28







Adddison

















asked Dec 19 '18 at 12:10









AdddisonAdddison

847




847












  • $begingroup$
    Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
    $endgroup$
    – lulu
    Dec 19 '18 at 12:13












  • $begingroup$
    If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
    $endgroup$
    – player100
    Dec 19 '18 at 12:24










  • $begingroup$
    It does not have to be exactly k,I added an example for clearance.@player100
    $endgroup$
    – Adddison
    Dec 19 '18 at 12:29


















  • $begingroup$
    Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
    $endgroup$
    – lulu
    Dec 19 '18 at 12:13












  • $begingroup$
    If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
    $endgroup$
    – player100
    Dec 19 '18 at 12:24










  • $begingroup$
    It does not have to be exactly k,I added an example for clearance.@player100
    $endgroup$
    – Adddison
    Dec 19 '18 at 12:29
















$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13






$begingroup$
Hint: use linearity. if $iin {1,cdots, n-(k-1)}$ then $i$ can be the start of a consecutive sequence of $k$ $1's$ so let $X_i$ be the indicator value of that event and work from there.
$endgroup$
– lulu
Dec 19 '18 at 12:13














$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24




$begingroup$
If there is a subsequence of $N_{k+2}$, for instance, does that contain 3 $N_k$ sequences, 1, or none? In other words does it have to be exactly k long to count?
$endgroup$
– player100
Dec 19 '18 at 12:24












$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29




$begingroup$
It does not have to be exactly k,I added an example for clearance.@player100
$endgroup$
– Adddison
Dec 19 '18 at 12:29










1 Answer
1






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oldest

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3












$begingroup$

$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
    $endgroup$
    – Adddison
    Dec 19 '18 at 16:21












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









3












$begingroup$

$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
    $endgroup$
    – Adddison
    Dec 19 '18 at 16:21
















3












$begingroup$

$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
    $endgroup$
    – Adddison
    Dec 19 '18 at 16:21














3












3








3





$begingroup$

$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.






share|cite|improve this answer









$endgroup$



$N_k=sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1}.$ Hence $E(N_k)=E(sum_{i=1}^{n-k+1}X_iX_{i+1}ldots X_{i+k-1})=(n-k+1)E(X_1,X_2ldots X_k)=frac{n-k+1}{2^k}$ if we have independence among $X_i$'s.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 19 '18 at 13:24









John_WickJohn_Wick

1,616111




1,616111












  • $begingroup$
    add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
    $endgroup$
    – Adddison
    Dec 19 '18 at 16:21


















  • $begingroup$
    add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
    $endgroup$
    – Adddison
    Dec 19 '18 at 16:21
















$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21




$begingroup$
add on to the question:What is the probability of not having a sequence of length $frac{1}{2}log_2n$?
$endgroup$
– Adddison
Dec 19 '18 at 16:21


















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